Survey
* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
List of important publications in mathematics wikipedia , lookup
History of mathematical notation wikipedia , lookup
Mathematics of radio engineering wikipedia , lookup
Line (geometry) wikipedia , lookup
Analytical mechanics wikipedia , lookup
Recurrence relation wikipedia , lookup
Elementary algebra wikipedia , lookup
System of polynomial equations wikipedia , lookup
Routhian mechanics wikipedia , lookup
System of linear equations wikipedia , lookup
KS3 Mathematics A2 Equations 1 of 62 © Boardworks Ltd 2004 Contents A2 Equations A A2.1 Solving simple equations A A2.2 Equations with the unknown on both sides A A2.3 Solving more difficult equations A A2.4 Equations and proportion A A2.5 Non-linear equations 2 of 62 © Boardworks Ltd 2004 Equations An equation links an algebraic expression and a number, or two algebraic expressions with an equals sign. For example, x + 7 = 13 is an equation. In an equation the unknown usually has a particular value. Finding the value of the unknown is called solving the equation. x + 7 = 13 x=6 When we solve an equation we always line up the equals signs. 3 of 62 © Boardworks Ltd 2004 Using inverse operations In algebra, letter symbols represent numbers. Rules that apply to numbers in arithmetic apply to letter symbols in algebra. For example, in arithmetic, if 3 + 7 = 10, we can use inverse operations to write: 10 – 7 = 3 and 10 – 3 = 7 In algebra, if a + b = 10, we can use inverse operations to write: or 4 of 62 10 – b = a and 10 – a = b a = 10 – b and b = 10 – a © Boardworks Ltd 2004 Using inverse operations In arithmetic, if 3 × 4 = 12, we can use inverse operations to write: 12 ÷ 4 = 3 and 12 ÷ 3 = 4 In algebra, if ab = 12, we can use inverse operations to write: 12 = a b or 5 of 62 a = 12 b and 12 = b a and b = 12 a © Boardworks Ltd 2004 Using inverse operations to solve equations We can use inverse operations to solve simple equations. For example, x + 5 = 13 x = 13 – 5 x=8 Always check the solution to an equation by substituting the solution back into the original equation. If we substitute x = 8 back into x + 5 = 13 we have 8 + 5 = 13 6 of 62 © Boardworks Ltd 2004 Using inverse operations to solve equations Solve the following equations using inverse operations. 17 – x = 6 5x = 45 x = 45 ÷ 5 x=9 Check: 5 × 9 = 45 17 = 6 + x 17 – 6 = x We always write 11 = x the letter before x = 11 the equals sign. Check: 17 – 11 = 6 7 of 62 © Boardworks Ltd 2004 Using inverse operations to solve equations Solve the following equations using inverse operations. x 7 =3 x=3×7 x = 21 Check: 21 7 =3 3x – 4 = 14 3x = 14 + 4 3x = 18 x = 18 ÷ 3 x=6 Check: 3 × 6 – 4 = 14 8 of 62 © Boardworks Ltd 2004 What number am I thinking of…? I’m thinking of a number. When I multiply the number by 2 and add 5 the answer is 11. What number am I thinking of …? We can write this as an equation. Instead of using ? for the number I am thinking of, let’s use the letter x. Start with x Multiply by 2 and add 5 x 2x 2x + 5 9 of 62 to give you 11. 2x + 5 = 11 © Boardworks Ltd 2004 What number am I thinking of…? Start with x Multiply by 2 and add 5 x 2x 2x + 5 to give you 11. 2x + 5 = 11 We can solve this equation using inverse operations in the opposite order. Start with the equation: 2x + 5 = 11 Subtract 5: 2x = 11 – 5 2x = 6 Divide by 2: x=6÷2 x=3 10 of 62 © Boardworks Ltd 2004 Solving equations by transforming both sides Solve this equation by transforming both sides in the same way: m –1=2 4 +1 +1 Add 1 to both sides. m =3 4 ×4 ×4 Multiply both sides by 4. m = 12 We can check the solution by substituting it back into the original equation: 12 ÷ 4 – 1 = 2 11 of 62 © Boardworks Ltd 2004 Find the missing number The number in each brick is found by adding the two numbers above it. What are the missing values? 23 ? n 23?+ n 62 n +?62 127 We can start by calling the unknown number in the top brick n. The unknown numbers in the next two bricks can be written in terms of n. We can now write an equation and solve it to find the value of n. 12 of 62 © Boardworks Ltd 2004 Find the missing number The number in each brick is found by adding the two numbers above it. What are the missing values? 23 n 21 2344 +n 62 n+ 8362 127 23 + n + n + 62 = 127 Simplify: 85 + 2n = 127 Subtract 85: 2n = 42 Divide by 2: n = 21 Check this solution. 13 of 62 © Boardworks Ltd 2004 Contents A2 Equations A A2.1 Solving simple equations A A2.2 Equations with the unknown on both sides A A2.3 Solving more difficult equations A A2.4 Equations and proportion A A2.5 Non-linear equations 14 of 62 © Boardworks Ltd 2004 Constructing an equation Ben and Lucy have the same number of sweets. Ben started with 3 packets of sweets and ate 11 sweets. Lucy started with 2 packets of sweets and ate 3 sweets. How many sweets are there in a packet? Let’s call the number of sweets in a packet, n. We can solve this problem by writing the equation: 3n – 11 = 2n – 3 The number of Ben’s sweets 15 of 62 is the same as the number of Lucy’s sweets. © Boardworks Ltd 2004 Solving the equation Let’s solve this equation by transforming both sides of the equation in the same way. 3n – 11 = 2n – 3 Start by writing the equation down. -2n Subtract 2n from both sides. -2n n – 11 = –3 +11 +11 n = 8 Always line up the equals signs. Add 11 to both sides. This is the solution. We can check the solution by substituting it back into the original equation: 3 8 – 11 = 2 8 – 3 16 of 62 © Boardworks Ltd 2004 Constructing an equation I’m thinking of a number. When I multiply the number by 4, I get the same answer as adding 9 to the number. What number am I thinking of? Let’s call the unknown number n. We can solve this problem by writing the equation: 4n The number multiplied by 4 17 of 62 = n+9 is the same as the number plus 9. © Boardworks Ltd 2004 Solving the equation Let’s solve this equation by transforming both sides of the equation in the same way. 4n = n + 9 -n -n Start by writing the equation down. Subtract n from both sides. 3n = 9 Always line up the equals signs. ÷3 Divide both sides by 3. ÷3 n = 3 This is the solution. We can check the solution by substituting it back into the original equation: 43= 3+9 18 of 62 © Boardworks Ltd 2004 Constructing an equation Find the value of x. (65 – 2x)o (2x + 5)o Remember, vertically opposite angles are equal. We can solve this problem by writing the equation: 2x + 5 = 65 – 2x 19 of 62 © Boardworks Ltd 2004 Solving the equation Let’s solve this equation by transforming both sides of the equation in the same way. 2x + 5 = 65 – 2x +2x +2x Add 2x to both sides. 4x + 5 = 65 -5 4x -5 = ÷4 x 60 ÷4 = Subtract 5 from both sides. 15 Divide both sides by 4. This is the solution. Check: 2 15 + 5 = 65 – 2 15 20 of 62 © Boardworks Ltd 2004 Rectangle problem The area of this rectangle is 27 cm2. Calculate the value of x and use it to find the height of the rectangle. 8x – 14 2x + 1 Opposite sides of a rectangle are equal. We can use this fact to write an equation in terms of x. 21 of 62 © Boardworks Ltd 2004 Rectangle problem The area of this rectangle is 27 cm2. 8x – 14 = 2x + 1 –2x –2x 6x – 14 = 1 8x – 14 2x + 1 +14 +14 6x = 15 ÷6 ÷6 x = 2.5 If x = 2.5 we can find the height of the rectangle using substitution: 8 × 2.5 – 14 = 20 – 14 = 6 cm 22 of 62 © Boardworks Ltd 2004 Rectangle problem The area of this rectangle is 27 cm2. What is its width? y Let’s call the width of the rectangle y. 8x – 14 2x + 1 The dimensions of the rectangle are 6 cm by 4.5 cm. If the height of the rectangle is 6 cm and the area is 27 cm2 then we can find the width by writing the equation: 6y = 27 ÷6 ÷6 y = 4.5 23 of 62 © Boardworks Ltd 2004 Select the correct equation Veronica has 58p and buys 4 chocolate bars. Thomas has £1 and buys 7 chocolate bars. They both receive the same amount of change. If c is the cost of one chocolate bar, which equation could we use to solve this problem? 24 of 62 A: 4c + 58 = 7c + 100 B: 58 – 4c = 1 – 7c C: 58 – 4c = 100 – 7c D: 4c – 58 = 7c – 1 © Boardworks Ltd 2004 Contents A2 Equations A A2.1 Solving simple equations A A2.2 Equations with the unknown on both sides A A2.3 Solving more difficult equations A A2.4 Equations and proportion A A2.5 Non-linear equations 25 of 62 © Boardworks Ltd 2004 Equations with brackets Equations can contain brackets. For example: 2(3x – 5) = 4x To solve this we can Multiply out the brackets: 6x –10 = 4x + 10 Add 10 to both sides: 6x = 4x + 10 - 4x Subtract 4x from both sides: 26 of 62 - 4x 2x = 10 ÷2 Divide both sides by 2: + 10 ÷2 x=5 © Boardworks Ltd 2004 Equations with brackets Sometimes we can solve equations such as: 2(3x – 5) = 4x by first dividing both sides by the number in front of the bracket: Divide both sides by 2: 3x – 5 = 2x +5 Add 5 to both sides: 3x = 2x + 5 - 2x Subtract 2x from both sides: +5 - 2x x=5 In this example, dividing first means that there are fewer steps. 27 of 62 © Boardworks Ltd 2004 Solving equations involving division Linear equations with unknowns on both sides can also involve division. 3x + 2 = 11 – x For example, 4 In this case we must start by multiplying both sides of the equation by 4. 3x + 2 = 4(11 – x) Multiply out the brackets: 3x + 2 = 44 – 4x Add 4x to both sides: 7x + 2 = 44 Subtract 2 from both sides: Divide both sides by 7: 28 of 62 7x = 42 x=6 © Boardworks Ltd 2004 Solving equations involving division Sometimes the expressions on both sides of the equation are divided. 4 5 For example, = (x + 3) (3x – 5) In this example, we can multiply both sides by (x + 3) and (3x – 5) in one step to give: 4(3x – 5) = 5(x + 3) Multiply out the brackets: Subtract 5x from both sides: Add 20 to both sides: Divide both sides by 7: 29 of 62 12x – 20 = 5x + 15 7x – 20 = 15 7x = 35 x=5 © Boardworks Ltd 2004 Contents A2 Equations A A2.1 Solving simple equations A A2.2 Equations with the unknown on both sides A A2.3 Solving more difficult equations A A2.4 Equations and proportion A A2.5 Non-linear equations 30 of 62 © Boardworks Ltd 2004 Sale! Footballs! Were £4, Now only £3! Original Price Sale Price We can write the ratio of 3 4 = 6 8 = 9 12 £4 £3 £8 £6 sale price original price = 12 16 £12 £16 £20 £9 £12 £15 for each pair of values. = 15 20 = 0.75 What do you The notice ratios are about equal. these ratios? 31 of 62 © Boardworks Ltd 2004 Drawing a graph Original Price, x £4 Sale Price, y £3 £8 £6 £12 £16 £20 £9 £12 £15 y Sale price of footballs 15 The points lie on a straight line through the origin. Sale price, £ 12 The equation of the line is: 9 3 6 y= 4 x or 3 0 0 4 8 12 16 20 x y = 0.75x Original price, £ 32 of 62 © Boardworks Ltd 2004 Constant speed Susan walks to school at a constant speed. Altogether, it takes her 10 minutes to walk 800 metres. How far would she have walked in two minutes? 2 4 6 8 Time, minutes 10 Distance, metres 160 320 480 640 800 We can write the ratio of 160 2 = 320 4 = distance 480 6 time = for each pair of values. 640 8 = 800 10 = 80 What do you The notice ratios are about equal. these ratios? The time and the distance are directly proportional. 33 of 62 © Boardworks Ltd 2004 Distance/time graph We can plot the points from the table onto a graph. Distance, metres Time, minutes 2 4 6 8 10 Distance, metres 160 320 480 640 800 y 800 Susan’s walking speed The points lie on a straight line through the origin. 640 The equation of the line is: y = 80x 480 320 160 0 0 2 4 6 8 10 x How far would Susan walk in: a) 3 minutes? 240 metres b) 15 minutes? 1200 metres Time, minutes 34 of 62 © Boardworks Ltd 2004 Mixing colours Orange paint is made by mixing 4 litres of yellow paint with 6 litres of red paint. 1 l 1 l 1 l 1 l 1 l 1 l 1 l 1 l 1 l 1 l = 10 l of orange paint To make the same shade of orange we must keep the amount of yellow paint and red paint in direct proportion. How many litres of each colour do you need to make: a) 5 litres of orange paint? 2 l of yellow and 3 l of red b) 1 litre of orange paint? 0.4 l of yellow and 0.6 l of red c) 7 litres of orange paint? 2.8 l of yellow and 4.2 l of red 35 of 62 © Boardworks Ltd 2004 Mixing colours 1 l 1 l 1 l 1 l 1 l 1 l 1 l 1 l 1 l 1 l = 10 l of orange paint The ratio red yellow is 6 4 = 1.5 We can write an equation linking the amount of red paint r to the amount of yellow paint y as r = 1.5y How many litres of red paint would be needed to mix with 14 litres of yellow paint to make the same shade of orange? r = 1.5 × 14 = 21 l 36 of 62 © Boardworks Ltd 2004